The first order reaction `2N_(2)O(g)rarr2N_(2)(g)+O_(2)(g)` has a rate constant of `1.3 xx 10^(-11)s^(-1)` at `270^(@)C` and `4.5 xx 10^(-10)s^(-1)` at `350^(@)C`. What is the activation energy for this reaction ?

Options

A. 15 kJ
B. 30 kJ
C. 68 kJ
D. 120 kJ

Explanation

To find the activation energy (Ea) for the given first-order reaction, we will use the Arrhenius equation in its logarithmic form:

\[ \log \left( \frac{K_2}{K_1} \right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

### Step 1: Convert temperatures from Celsius to Kelvin
- \( T_1 = 270^\circ C + 273 = 543 \, K \)
- \( T_2 = 350^\circ C + 273 = 623 \, K \)

### Step 2: Identify the rate constants
- \( K_1 = 1.3 \times 10^{-11} \, s^{-1} \)
- \( K_2 = 4.5 \times 10^{-10} \, s^{-1} \)

### Step 3: Substitute the values into the Arrhenius equation
We can now substitute the values into the equation:

\[ \log \left( \frac{K_2}{K_1} \right) = \log \left( \frac{4.5 \times 10^{-10}}{1.3 \times 10^{-11}} \right) \]

Calculating the logarithm:

\[ \frac{K_2}{K_1} = \frac{4.5 \times 10^{-10}}{1.3 \times 10^{-11}} \approx 34.6154 \]

Now, calculate the logarithm:

\[ \log(34.6154) \approx 1.539 \]

### Step 4: Calculate the difference in the reciprocal of temperatures
Now we calculate:

\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{543} - \frac{1}{623} \]

Calculating each term:

\[ \frac{1}{543} \approx 0.001838 \, K^{-1} \]
\[ \frac{1}{623} \approx 0.001606 \, K^{-1} \]

Now, find the difference:

\[ 0.001838 - 0.001606 \approx 0.000232 \, K^{-1} \]

### Step 5: Substitute into the Arrhenius equation
Now we substitute everything back into the logarithmic form of the Arrhenius equation:

\[ 1.539 = \frac{E_a}{2.303 \times 8.314} (0.000232) \]

### Step 6: Solve for \(E_a\)
Rearranging gives:

\[ E_a = \frac{1.539 \times 2.303 \times 8.314}{0.000232} \]

Calculating the right-hand side:

\[ E_a \approx \frac{1.539 \times 2.303 \times 8.314}{0.000232} \approx 124606 \, J/mol \]

### Step 7: Convert to kJ/mol
To convert Joules to kilojoules:

\[ E_a \approx 124.606 \, kJ/mol \]

### Conclusion
Thus, the activation energy for the reaction is approximately:

\[ E_a \approx 124 \, kJ/mol \]